I have an 80 g·cm motor with a rotational frequency of 15,000 rpm. I want to lift a weight of 2 kg at a speed of 0.5 m/s. How do I calculate the gear ratio required?
I have a 80gcm motor with a rpm of 15000.
I want to lift a weight of 2 kg at a speed of 0.5 m/s.
How do I go about calculating the gear ratio required for this?
Firstly - is it possible?
In particular, is there enough input power available for the desired output power?
To within about 2% a very handy formula applies - it can be derived conventionally and seen that several factors happen to cancel nicely.
Watts = kg x metres x RPM
80 gram∙cm = 0.080 kg x 0.01 m
So for input W = 0.080 kg x 0.01 m x 15000 = 12 Watts.
This is the maximum Wattage you can deliver if properly geared at 100% efficiency
(we should be so lucky).
Desired power = Force x distance per unit time
Watts = Joules/sec = mg∙d/s
= 2 kg x $g$ x 0.5 m/s = 2 x 9.8 x 0.5 = 9.8 Watts
So to work at all overall efficiency needs to be at least 9.8/12
or greater than about 82%.
That's potentially doable but also potentially difficult.
Now to the actual problem.
The following assumes that the output weight or force is taken from the end of a radius of the driven "gear". If output is instead taken from eg a windlass drum at lower diameter to the driven gear the ratios will be scaled based on the relative diameters. Ignore that for now.
Torque_in x RPM_in = Torque_out x RPM_out at 100% efficiency
or RPM_out = Torque_in x RPM_in / Torque_out at 100% efficiency
RPM out = 0.080 kg x 0.01 m x 15000 RPM / (2 kg x 0.5 m) = 12 RPM
So gear ratio = 15000/12 = 1250:1
The specification of not just output Torque but actual force (2 kg x $g$) constrains the actual output pulley size if output is taken off at the pulley radius.